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lubasha [3.4K]
3 years ago
10

9) One line of text on a page uses about 3/16 of an inch. There are 1-inch 10 margins at the top and bottom of a page. Write and

solve an inequality to find the number of lines that can be typed on a page that is 11 inches long.​
Mathematics
1 answer:
arsen [322]3 years ago
8 0
11-2(1)
This is for the top and bottom margins. You are left with 9. Divide that by 3/16 or 0.1875 to get 48.
(11-2)/0.1875
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Answer:

3q - 5 = 13 Q = 6

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a rectangular box with two sqaure opposite ends is to hold 8000 cubic inches. find the dimensions of the cheapest box if tge rec
AveGali [126]

Answer:

the dimensions of the box that minimizes the cost are 5 in x 40 in x 40 in

Step-by-step explanation:

since the box has a volume V

V= x*y*z = b=8000 in³

since y=z (square face)

V= x*y² = b=8000 in³

and the cost function is

cost = cost of the square faces *  area of square faces + cost of top and bottom * top and bottom areas + cost of the rectangular sides * area of the rectangular sides

C = a* 2*y² +  a* 2*x*y + 15*a* 2*x*y =  2*a* y² +  32*a*x*y

to find the optimum we can use Lagrange multipliers , then we have 3  simultaneous equations:

x*y*z = b

Cx - λ*Vx = 0 → 32*a*y -  λ*y² = 0 → y*( 32*a-λ*y) = 0 → y=32*a/λ

Cy - λ*Vy = 0  → (4*a*y + 32*a*x) - λ*2*x*y = 0

4*a*32/λ  + 32*a*x - λ*2*x*32*a/λ = 0

128*a² /λ +  32*a*x - 64*a*x = 0

32*a*x = 128*a² /λ

x  = 4*a/λ

x*y² = b

4*a/λ * (32*a/λ)² = b

(a/λ)³ *4096 =  8000 m³

(a/λ) = ∛ ( 8000 m³/4096 ) = 5/4 in

then

x  = 4*a/λ = 4*5/4 in = 5 in

y=32*a/λ = 32*5/4 in = 40 in

then the box has dimensions 5 in x 40 in x 40 in

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